Bonus excercise: use Monte Carlo to approximate a definite integral

library(tidyverse)
library(plotly)

Consider a somewhat nasty-looking curve in two dimensions. We define \(z\) to be a function of \(x\) and \(y\) that looks like this: \[ z = f(x, y) = \biggl[1 + e ^{-|x + 2y|} \cos\biggl(0.2 \sqrt{x^2 + y^2}\biggr) \biggr] \times \mathbb{I}\{-10 < x,y < 10 \} \] In other words, it is a function that returns 0 everywhere outside of a square of width 20, centered on the origin. But within that square it returns positive values.

Here is some code that defines it:

ugly <- function(x, y) {
  ifelse(x < -10 | y > 10 | x > 10 | y < -10, 
         0,
         1 + exp(-abs(x + 2 * y) / 10) * cos(0.2 * sqrt(x^2 * y^2))
  )
}

And, we can use that to visualize the surface by evaluating a lot of points and plotting it with the plotly package. Go ahead and rotate and manipulate the thing.

vals <- seq(-12, 12, by = 0.05)
pts <- expand_grid(
  x = vals,
  y = vals
) %>%
  mutate(z = ugly(x, y))

zmat <- matrix(pts$z, ncol = length(vals), nrow = length(vals))

plot_ly(x = vals, y = vals, z = ~zmat) %>%
  add_surface()

The volume under that curve is given by the definite integral: \[ \int_{-10}^{10} \int_{-10}^{10} f(x, y) dxdy = \int_{-10}^{10} \int_{-10}^{10} \biggl[1 + e ^{-|x + 2y|} \cos\biggl(0.2 \sqrt{x^2 + y^2}\biggr) \biggr] dxdy \] That looks a little hairy. It may have an analytical solution, or it could be evaluated numerically with high accuracy. But, if you rewrite that integral as an expection over a two-dimensional uniform variable on the square from -10 to 10 in \(x\) and also from -10 to 10 in \(y\), then you get: \[ 20 \times 20 \times \int_{-10}^{10} \int_{-10}^{10} f(x, y) \frac{1}{20}\frac{1}{20}dxdy = 400 \times \mathbb{E}[f(x, y)] \] where the expectation is taken over a 2-D uniform variable from -10 to 10 in both x and y.

Given that as a preamble, your mission it to approximate that integral using Monte Carlo sampling.

Good luck! The answer is \(\approx 423.7\).